Three Congruent Circles by Reflection II: What is this about?
A Mathematical Droodle
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Copyright © 1996-2018 Alexander BogomolnyThe applet attempts to illustrate a theorem by Quang Tuan Bui:
In ΔABC, L is the incenter, A_{b} and A_{c} are reflections of A in the angle bisectors BL and CL. B_{a}, B_{c}, C_{a}, and C_{b} are defined similarly. Then the circumcircles of triangles AC_{b}B_{c}, BA_{c}C_{a}, and CB_{a}A_{b} are congruent. |
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Proof
Reflect B_{c} and C_{b} in AL to obtain B'_{c} and C'_{b}. Similarly construct A'_{c}, A'_{b}, C'_{a} and B'_{a}. Observe that (as vectors)
BC'_{a} = CB'_{a}, CA'_{b} = AC'_{b}, AB'_{c} = BA'_{c}. |
If O is the circumcenter of ΔABC we can define points X, Y, Z so that
OX = BC'_{a} = CB'_{a}, OY = CA'_{b} = AC'_{b}, OZ = AB'_{c} = BA'_{c}. |
Thus we also obtain three triangle congruencies:
ΔOXY = ΔCB'_{a}A'_{b} ΔOYZ = ΔAC'_{b}B'_{c}, ΔOZX = ΔA'_{c}C'_{a}. |
(In fact the triangles are not just congruent, they are obtained from each other by translations with vectors OC, OB, and OA, respectively.) We need to prove that the three triangles on the left have equal circumcircles. To this end, suffice it to show that the four points O, X, Y, Z are concyclic.
By construction, say,
LC = LC_{a} = LC'_{a} |
so that LC = LC'_{a} making the projection of L onto BC the midpoint of CC'_{a} and also the midpoint of BB'_{a}. This implies that the trapezoid OXC'_{a}C is isosceles, with L on the orthogonal bisectors of one of the bases (CC'_{a}). It then is also on the orthogonal bisector of the other base, viz., OX, implying
LO = LX. |
Similarly,
LO = LY and IO = LZ. |
The points O, X, Y, Z are indeed concyclic which prove our statement: the three (actually 6) circles at hand are congruent to L(O), the circle through O with center L.
As an extra, we'll show that the centers of the six circles all lie on the circle with center L and radius R, the circumradius of ΔABC.
Indeed, let O_{a}, O_{b}, O_{c}, O'_{a}, O'_{b}, O'_{c} be the circumcenters of triangles AB_{c}C_{b}, BA_{c}C_{a}, CB_{a}A_{b}, AB'_{c}C'_{b}, BA'_{c}C'_{a}, CB'_{a}A'_{b}. Since primed triangles are translations of those in L(O), all segments AO'_{a}, BO'_{b}, and CO'_{c} are equal and parallel to OL giving
LO'_{a} = LO'_{b} = LO'_{c} = R. |
But O_{a}, O_{b}, O_{c} are reflections of O'_{a}, O'_{b}, O'_{c} in the respective angle bisectors, from which also
LO_{a} = LO_{b} = LO_{c} = R, |
and we are done.
An additional fact is worth mentioning: since AO'_{a}, BO'_{b}, and CO'_{c} are parallel, they can be said to concur at infinity. Their isogonal reflections AO_{a}, BO_{b}, and CO_{c}
also concur with the point of concurrence lying on the circumcircle of ΔABC.References
- Quang Tuan Bui, Two Triads of Congruent Circles from Reflections, Forum Geometricorum, Volume 8 (2008) 7–12.
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Copyright © 1996-2018 Alexander Bogomolny